# Strength of material in hollow shaft torque and safety factor

1. Jan 20, 2016

### lonlyway

1. The problem statement, all variables and given/known data

the tube has ID of 450mm, t= 6mm, OD= 462mm, pressure is 1.2MPa
This is based on ASTM A36 steel, which have
Ultimate Tensile strength 400-550MPa
Yield Tensile strength 250Mpa
Modululs of Elasticity 200Gpa
Shear Modulus 79.3GPa
Determine the factor of safety at points H and K along the top of the tank and point G on the side by using a) Maximum shear stress theory, b) D.E Von Mises theory

2. Relevant equations

3. The attempt at a solution
What I am thinking is using pure tortion on the AD axle, and get T value
T = 0.5m x 5000N = 2500Nm or 2.5KNm
than find the moment of inertia by using J = pi(OD4-ID4)/2 = 7.1504 x 10-3 m4
at this point, I have G=79.3 x 109, J = 7.1504 x 10-3 m4, and need to find τ by using τ=pT/J = 0.231 x 2500/(7.1504 x 10-3) = 80764.71246 N/m2 which tives τxy on point H K and G
Than thinking to find stress of hoop and axial, by using
σhoop = pr/t and σaxial = pr/2t on surface, than calculate all and convert into σL and σG, H or K
which will give me max stress, than devide the max stress by the σultimate which will give me safety factor

Am I on the right track? if not, what did I do wrong?
and how should I use D.E Von Mises theory on the question?

Last edited: Jan 20, 2016
2. Jan 20, 2016

### SteamKing

Staff Emeritus
Your formula for J is incorrect. For a circular tube, J = (π/2) ⋅ (ro4 - ri4 )

The manner in which the torque is applied to this tube also leads to some bending stress. You should also look at this tube as a cantilever beam, given the 5 kN load applied at Point D.

3. Jan 20, 2016

### lonlyway

Oops yeah it was radius not diameter. I will fix it. Thank you
I did not put any bending moment as cantilever since there was no dimension about the small bar at D such as diameter. Therefore I assume that in this question, the force transfer without any loss. - which I am not really sure its correct assumption

4. Jan 20, 2016

### SteamKing

Staff Emeritus
You don't care about the small bar sticking out the side of the tube.

The force can be transferred back to the center of the tube along with a couple which creates the torque on the tube. Applying a force away from the point where the tube is fixed to the wall is going to create a bending moment and bending stresses at points H and K. You'll also have to check point G for shearing stress due to the vertical load, which is why I mentioned that the internal pressure and the torsional loads are not the only loads to be considered here, especially since you are supposed to be checking for factor of safety.

5. Jan 20, 2016

### lonlyway

Oh okay, thanks completely missed that.
So now I have 3 forces, bending moment as cantilever beam or shear stress depends on location, torsion, and hoof and axial stress and combine all 3 have 2 prime stresses per locations.
I can get max shear theory easily, but what should I do for Von Mises?
I have checked wikipedia, but it confuses me.
Should I just assume σ3 = 0 and sub σ1(length) and σ(hoof) into this equation?

6. Jan 20, 2016

### SteamKing

Staff Emeritus
The stresses σ1, σ2, and σ3 are principal stresses, so you can't assume that one is equal to zero. You'll have to work out these principal stresses after analyzing the tube.